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Is there a simple description of the fundamental group of algebraic complex curves, possibly in terms of generators and relations? I am especially interested in the case of singular curves, as the correspondence with Riemann surfaces answers the question in the smooth case and gives an answer depending only on the genus of the curve. As a follow-up, I would also like to know what happens if one removes a finite set of points from the curve, which is also relatively easy to answer for Riemann surfaces.

Edit: More precisely: The ordinary fundamental group of the underlying topological space of the curve embedded into either the affine or the projective space.

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  • $\begingroup$ Are you looking for algebraic varieties? Otherwise, I'm not sure. $\endgroup$ – Don Thousand Jul 15 '19 at 12:11
  • $\begingroup$ Yes, that might have been unclear. I am interested in algebraic varieties of dimension 1 over the complex numbers. $\endgroup$ – Benjamin Zayton Jul 15 '19 at 12:25
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    $\begingroup$ Okay, that makes a lot more sense. Curves can mean a whole variety of things (pun not intended) $\endgroup$ – Don Thousand Jul 15 '19 at 12:26
  • $\begingroup$ And are you simply talking about the fundamental group of the underlying topological space of the curve? With the subspace topology induced from, say, embedding it into $\mathbb CP^n$? Where $\mathbb CP^n$ is endowed with its topology as an $n-1$ dimensional complex manifold? $\endgroup$ – Lee Mosher Jul 15 '19 at 14:15
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    $\begingroup$ I'm trying to make sure that you are not talking about any kind of special algebraic geometry analogue of the ordinary topological notion of fundamental group; if you are talking about something like that, I have nothing to offer. $\endgroup$ – Lee Mosher Jul 15 '19 at 14:16
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I'll take a crack at this, although my knowledge of algebraic geometry is extremely slim. I'll stick to the projective case at first, where the curve is compact as a topological space. Then I'll add comments about the affine case.

I think that the following statement is true, and I shall hang my answer on it: if $X$ is a complex projective curve, and if $X$ is connected as a topological space, then $X$ is homeomorphic to a quotient of a closed, oriented surface $S$ where the quotient map $q : S \mapsto X$ is finite-to-one and $q$ is one-to-one over the complement of a finite subset of $X$. The idea here is that there might be "cusp" singularities where $X$ is locally a 2-dimensional topological manifold (i.e. locally homeomorphic to $\mathbb C \approx \mathbb R^2$), and those points are okay. But near "node" singularities, one must pull some stuff apart.

Assuming this is correct, it follows that $\pi_1(X)$ is the free product of several terms. Each of these terms -- each free factor -- is a group whose presentation is known. You can combine the presentations of the free factors into a single presentation for $\pi_1(X)$.

Here are the free factors, in outline. First, each component of $S$ is a compact connected Riemann surface of some genus $g$, and the fundamental group of that component is a free factor; a presentation for this free factor is known, expressed in terms of $g$. There is one more term, namely a certain finite rank free group $F_X$; its presentation is also known, expressed in terms of the rank.

So all I really have to do now is to explain how to calculate the rank of the free factor $F_X$. You can do this easily from the pattern of identifications of $q$. I'll give an inductive procedure for determining the rank, which allows one to inductively apply a very simple form of Van Kampen's Theorem. Then I'll write down a closed form for the rank of $F_X$.

Let's break the quotient $q : S \to X$ into a series of small steps. In each step we identify a two point subset to a single point. The result is a factorization of $q$ into a sequence of quotient maps $$S = S_0 \mapsto S_1 \mapsto ... \mapsto S_K = X $$ Each quotient map $S_{k-1} \mapsto S_k$ identifies a two point subset to a single point. I'll define a sequence of free groups $F_0,F_1,...F_K$ of non-strictly increasing ranks, starting with $F_0$ which is a rank $0$ free group (a trivial group). Inductively, for $k=1,...,K$, consider the two subset of $S_{k-1}$ that is identified under the quotient map $S_{k-1} \mapsto S_k$. If those two points are contained in the same component of $S_{k-1}$ then $F_k$ is a free group of rank one larger than the rank of $F_{k-1}$; otherwise, $F_k$ is a free group of the same rank as that of $F_{k-1}$. Finally, let $F_X = F_K$.

One can deduce a closed form from all of this: if the number of components of $S$ equals $C$, and if the non-one-to-one points of $q$ have preimages of cardinalities $D_1,...,D_I$, then the rank of $F_X$ is equal to $$1 - C + \sum_{i=1}^I (D_i-1) = 1 - (I + C) + \sum_{i=1}^I D_i $$

The affine case, and the case where you remove a finite set of points, can be handled in the same fashion, except for the following alteration: a component of $S$ is allowed to be a noncompact oriented surface of finite type, obtained from a surface of genus $g \ge 0$ by removing some number $b \ge 1$ of its points, in which case the fundamental group of that component is itself a free group of rank $2g+b-1$.

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  • $\begingroup$ Could you help me with this please math.stackexchange.com/questions/3292324/…? Thank you! $\endgroup$ – Nash Jul 15 '19 at 19:31
  • $\begingroup$ After a bit of research, your central assumption certainly appears to be true, because every irreducible curve $C$ in complex projective space admits a normalization, i.e. a holomorphic map $\sigma:X \to \mathbb{C}\mathbb{P}^2$, where $X$ is a compact Riemann surface, such that if $S$ denotes the singular points on $C$, $\sigma(X)=C$, $\sigma^-1(S)$ is a finite set, and the restriction $\sigma: X\setminus \sigma^-1(S) \to C\setminus S$ is bijective. (Griffiths' Introduction to algebraic curves is a reference). Thank you! $\endgroup$ – Benjamin Zayton Jul 16 '19 at 15:54
  • $\begingroup$ Ah hah! I could not have said all of those words. $\endgroup$ – Lee Mosher Jul 16 '19 at 17:46

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